Question

Calculate the pH buffer composed of 0.12 mol benzoic acid and 0.20 mol of sodium benzoate? Ka = 6.3X10 ^ -5.

1) Calculate the pH after 0.020 mol of HCl have been added

2) Calculate the pH after 0.020 mol of NaOH have been added

Answer 1

pH= 4.42; 4.31; 4.54

Explanation:

pH of a buffer solution
`pH= pKa+ log ((Cs)/(Ca)) = 4,20 +log(0.2/0.12) =4.42`

if you add 0,02 mol of HCl this reacts with sodium benzoate decreasing of 0,02 mol its quantity (that will be 0,18) and increasing of 0,02 mol the quantity of benzoic acid (that will be 0,14) so you will have
`pH= pKa+ log ((Cs)/(Ca)) = 4,20 +log(0.18/0.14) =4.31`

if you add 0,02 mol of NaoH this reacts with benzoic acid decreasing of 0,02 mol its quantity (that will be 0,10) and increasing of 0,02 mol the quantity of sodium benzoate (that will be 0,22) so you will have
`pH= pKa+ log ((Cs)/(Ca)) = 4,20 +log(0.22/0.10) =4.54`